To answer this question, we first need to understand the structure of a cache memory address. A cache memory address is divided into three parts: the Tag, the Set, and the Word. The number of bits in each part depends on the configuration of the cache memory and the memory unit of the computer.

1. First, let's calculate the number of bits needed for the Word part. Since the cache memory has 1k words, it means there are 1024 words. To address 1024 different locations, we need log2(1024) = 10 bits.

2. Next, let's calculate the number of bits needed for the Set part. The memory unit of the computer is 64k x 16, which means there are 64k = 65536 words in the memory. Since each word in the memory can be stored in the cache, the number of sets in the cache is the total number of words in the memory divided by the number of words in the cache, which is 65536/1024 = 64 sets. To address 64 different sets, we need log2(64) = 6 bits.

3. Finally, the remaining bits are used for the Tag part. The total number of bits in the address format is the number of bits needed to address all the words in the memory, which is log2(65536) = 16 bits. So, the number of bits in the Tag part is the total number of bits in the address format minus the number of bits in the Set part and the Word part, which is 16 - 6 - 10 = 0 bits.

So, there are 0 bits in the Tag of the address format.

Question

To answer this question, we first need to understand the structure of a cache memory address. A cache memory address is divided into three parts: the Tag, the Set, and the Word. The number of bits in each part depends on the configuration of the cache memory and the memory unit of the computer.

1. First, let's calculate the number of bits needed for the Word part. Since the cache memory has 1k words, it means there are 1024 words. To address 1024 different locations, we need log2(1024) = 10 bits.

2. Next, let's calculate the number of bits needed for the Set part. The memory unit of the computer is 64k x 16, which means there are 64k = 65536 words in the memory. Since each word in the memory can be stored in the cache, the number of sets in the cache is the total number of words in the memory divided by the number of words in the cache, which is 65536/1024 = 64 sets. To address 64 different sets, we need log2(64) = 6 bits.

3. Finally, the remaining bits are used for the Tag part. The total number of bits in the address format is the number of bits needed to address all the words in the memory, which is log2(65536) = 16 bits. So, the number of bits in the Tag part is the total number of bits in the address format minus the number of bits in the Set part and the Word part, which is 16 - 6 - 10 = 0 bits.

So, there are 0 bits in the Tag of the address format.

Knowee AI · Accepted Answer

To answer this question, we first need to understand the structure of a cache memory address. A cache memory address is divided into three parts: the Tag, the Set, and the Word. The number of bits in each part depends on the configuration of the cache memory and the memory unit of the computer.

1. First, let's calculate the number of bits needed for the Word part. Since the cache memory has 1k words, it means there are 1024 words. To address 1024 different locations, we need log2(1024) = 10 bits.

2. Next, let's calculate the number of bits needed for the Set part. The memory unit of the computer is 64k x 16, which means there are 64k = 65536 words in the memory. Since each word in the memory can be stored in the cache, the number of sets in the cache is the total number of words in the memory divided by the number of words in the cache, which is 65536/1024 = 64 sets. To address 64 different sets, we need log2(64) = 6 bits.

3. Finally, the remaining bits are used for the Tag part. The total number of bits in the address format is the number of bits needed to address all the words in the memory, which is log2(65536) = 16 bits. So, the number of bits in the Tag part is the total number of bits in the address format minus the number of bits in the Set part and the Word part, which is 16 - 6 - 10 = 0 bits.

So, there are 0 bits in the Tag of the address format.

A digital computer space has a memory unit of 64k x 16 and a cache memory of 1k words. How manybits are there in the tag of the address format?*810166

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